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Chapter 14: Problem 13

Larry reads that one out of four eggs contains salmonella bacteria. So he never uses more than three eggs in cooking. If eggs do or don't contain salmonella independent of each other, the number of contaminated eggs when Larry uses three chosen at random has the distribution (a) binomial with \(n=4\) and \(p=1 / 4\). (b) binomial with \(n=3\) and \(p=1 / 4\). (c) binomial with \(n=3\) and \(p=1 / 3\).

Short Answer

Expert verified
(b) binomial with \( n=3 \) and \( p=1/4 \).

Step by step solution

01

Understand the Scenario

We are given that the probability of an egg containing salmonella is \( \frac{1}{4} \). Larry picks 3 eggs to use in cooking, and each egg is independent of the others in terms of salmonella presence.
02

Determine the Appropriate Probability Model

The problem requires finding out how many eggs could potentially be contaminated out of 3. This fits the binomial distribution, where we count the number of successes (contaminated eggs) in a fixed number of trials (3 eggs).
03

Set Parameters for Binomial Distribution

The binomial distribution is characterized by two parameters: \( n \) (the number of trials) and \( p \) (the probability of success on any given trial). Here, \( n = 3 \) (Larry only uses 3 eggs) and \( p = \frac{1}{4} \) (the probability of a single egg being contaminated).
04

Identify the Correct Answer Choice

With \( n = 3 \) and \( p = \frac{1}{4} \), the correct binomial distribution matching these parameters is (b) binomial with \( n=3 \) and \( p=1/4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Model
In statistics, a probability model is a mathematical representation of a random phenomenon. By using probability models, we can predict the likelihood of various outcomes. In the context of Larry's egg problem, the Random Variable "X" represents the number of contaminated eggs out of the ones used.

The probability model useful in this scenario is the **binomial distribution**. This distribution is suitable for problems where you have a fixed number of trials, each trial has only two possible outcomes (such as success or failure), and each trial is independent from the others. In Larry's case:
  • We are considering 3 trials (the three eggs he uses).
  • A "success" here is finding a contaminated egg.
  • The probability of success is 1 out of 4, or 0.25, for each egg.
By setting up a binomial distribution, we can calculate the probability of any given number of contaminated eggs among the ones Larry uses.
Independent Trials
The concept of independent trials is crucial when dealing with binomial distributions. Trials are independent if the outcome of one trial does not affect the outcome of another.
For example, imagine flipping a coin several times. Each flip is independent because the result of one flip does not influence others.

In Larry's scenario, **each egg has its own probability** of being contaminated. One egg being safe doesn鈥檛 change the chance that another is contaminated, making these trials independent.
This independence is what allows us to use the binomial distribution model properly. A lack of independence in trials could lead to different models or distributions being necessary since the basic assumption of no mutual influence is violated. However, in this problem, Larry's use of the eggs from a larger group ensures that the independence condition is met.
Success Probability
Success probability in a binomial setting refers to the chance of achieving the outcome of interest in each trial. In simpler terms, it's the probability that a single trial will result in a success.In the scenario given, the success probability is the likelihood that an egg is contaminated with salmonella. This is given as \( \frac{1}{4} \) or 0.25, meaning that each egg independently has a 25% chance of being contaminated.

The success probability affects the distribution and the expected outcomes. With a success probability of 0.25, the model expects low contamination among small batches of eggs. However, if Larry were to use, say, 100 eggs, we'd expect more contaminated eggs, as the probability accumulates over many trials. Understanding this probability aids in setting realistic expectations for outcomes in any real-world application using binomial distribution.

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Most popular questions from this chapter

Estimating \(\pi\) from random numbers. Kenyon College student Eric Newman used basic geometry to evaluate software random number generators as part of a summer research project. He generated 2000 independent random points \((X, Y)\) in the unit square. [That is, \(X\) and \(Y\) are independent random numbers between 0 and 1, each having the density function illustrated in Figure \(12.4\) (page 290). The probability that \((X, Y)\) falls in any region within the unit square is the area of the region. \(]^{14}\) (a) Sketch the unit square, the region of possible values for the point \((X, Y)\). (b) The set of points \((X, Y)\) where \(X^{2}+Y^{2}<1\) describes a circle of radius 1 . Add this circle to your sketch in part (a), and label the intersection of the two regions \(A\). (c) Let \(T\) be the total number of the 2000 points that fall into the region \(A\). \(T\) follows a binomial distribution. Identify \(n\) and \(p\). (Hint: Recall that the area of a circle is \(\pi r^{2}\)-) (d) What are the mean and standard deviation of \(T\) ? (e) Explain how Eric used a random number generator and the facts given here to estimate \(\pi\).

Using Benford's Law. According to Benford's law (Example 12.7, page 2B7) the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is \(0.477\). You examine 90 invoices from a vendor and find that 29 have first digits 1 or 2 . If Benford's law holds, the count of 1 s and \(2 s\) will have the binomial distribution with \(n=90\) and \(p=0.477\). Too few \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) suggests fraud. What is the approximate probability of 29 or fewer 15 and \(2 \mathrm{~s}\) if the invoices follow Benford's law? Do you suspect that the invoice amounts are not genuine?

Roulette-betting on red. A roulette wheel has 38 slots, numbered 0,00 , and 1 to 36 . The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. (a) If you bet on "red," you win if the ball lands in a red slot. What is the probability of winning with a bet on red in a single play of roulette? (b) You decide to play roulette four times, each time betting on red. What is the distribution of \(X\), the number of times you win? (c) If you bet the same amount on each play and win on exactly two of the four plays, then you will "break even." What is the probability that you will break even? (d) If you win on fewer than two of the four plays, then you will lose money. What is the probability that you will lose money?

Hiking the High Sierra loop in Yosemite. Yosemite National Park has five High Sierra camps arranged in a 49-mile loop. The camps are separated by 8-10 miles, which allows you to hike from one camp to the next in a day. Each camp provides tent cabins with four to six beds per cabin and serves family-style meals. Because of the popularity of the loop, you must enter a lottery in the fall for the following summer, and in a given year, approximately \(25 \%\) of the groups are selected at random to hike the loop." You plan to enter the lottery each year for the next five years. Let \(X\) be the number of years in which you are selected to hike the loop. (a) \(X\) has a binomial distribution. What are \(n\) and \(p\) ? (b) What are the possible values that \(X\) can take? (c) Find the probability of each value of \(X\). Draw a probability histogram for the distribution of \(\boldsymbol{X}\). (See Figure 14.2, page 339, for an example of a probability histogram-) (d) What are the mean and standard deviation of this distribution? Mark the location of the mean on your histogram.

If the subject guesses two shapes correctly and two incorrectly, in how many ways can you arrange the sequence of correct and incorrect guesses? (a) \((32)=3\left(\begin{array}{l}3 \\ 2\end{array}\right)=3\) (b) \((42)=12\left(\begin{array}{l}4 \\ 2\end{array}\right)=12\) (c) \((42)=6\left(\begin{array}{l}4 \\ 2\end{array}\right)=6\)
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